How many ammonium ions are in 25.0 picograms of ammonium biphosphite
To calculate the number of ammonium ions in 25.0 picograms of ammonium biphosphite (NH4H2PO3), we need to convert the given mass into moles and then use the Avogadro's number to determine the number of ions.
1. Calculate the molar mass of ammonium biphosphite (NH4H2PO3):
- Ammonium (NH4) has a molar mass of 14.01 + 4(1.01) = 18.05 g/mol.
- Biphosphite (H2PO3) has a molar mass of 2(1.01) + 31.00 + 3(16.00) = 97.03 g/mol.
- Adding both together, the molar mass of ammonium biphosphite is 18.05 + 97.03 = 115.08 g/mol.
2. Convert picograms (pg) to grams (g):
- 1 picogram (pg) is equal to 1 × 10^-12 grams (g).
- Therefore, 25.0 picograms is equal to 25.0 × 10^-12 grams or 2.5 × 10^-11 grams.
3. Convert grams to moles:
- Divide the given mass (in grams) by the molar mass of the compound: 2.5 × 10^-11 g / 115.08 g/mol = 2.17 × 10^-13 mol.
4. Determine the number of ammonium ions:
- The compound ammonium biphosphite (NH4H2PO3) contains 1 ammonium ion (NH4+) per molecule.
- Therefore, the number of ammonium ions in 2.17 × 10^-13 mol of ammonium biphosphite is also 2.17 × 10^-13 mol.
In conclusion, there are approximately 2.17 × 10^-13 ammonium ions in 25.0 picograms of ammonium biphosphite.